The moist weight of 0.19 ft3 of soil is 21.6 lb. If the moisture content is 18 percent and the specific gravity of soil solids is 2.55, find the following: (a) bulk density, (b) dry density, (c) void ratio, (d) porosity, (e) degree of saturation, and (f) the volume occupied by water (ft3).
What will be an ideal response?
(a) Solve for the bulk density, using Equation 17.9
? = W / V
? = 21.6 lb / 0.19 ft3
? = 114 lb/ft3
(b) Solve for the dry density, using Equation 17.11.
?d = ? / (1 + w)
?d = 113.68 / (1 + 0.18)
?d = 96 lb/ft3
(c) Solve for void ratio, e, using Equation 17.13.
e = [((1 + w) (Gs)(?w)) / ?] – 1
e = (((1 + 0.18) (2.55) (62.4)) / 113.68) – 1
e = 0.652
(d) Solve for porosity, n, using Equation 17.4.
n = e / (1 + e)
n = 0.652 / (1 + 0.652)
n = 0.395
(e) Compute the degree of saturation, S.
S = Vw / Vv = (w)(Gs) / e
S = (0.18)(2.55) / 0.652
S = 0.704 = 70.4%
(f) Determine the volume occupied by water, using Equation 17.7.
w = Ww / Ws
Ww = (w/(1 + w)) W
Ww = (0.18/1.18)(21.6) = 3.295 lb
Vw = Ww / ?w = (3.295 lb)/(62.4 lb/ft3)
Vw = 0.053 ft3
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